Answer:
= 9.8°
Explanation:
Width of one slit (a₁ ) = 1 / 1000 mm=0.001 mm = 10⁻⁶ m.
width of one slit in case 2 (a₂ ) = 1/500 =2 x 10⁻⁶ m
angular position of fringe, Sinθ = n λ /a
n is order of fringe , λ is wave length of light and a is slit aperture
So Sinθ ∝ 1 / a
Sin θ₁ /Sin θ₂ = a₂/a₁ ;
Sin20°/sinθ₂ = 2 / 1
sinθ₂ = Sin 20° / 2 = .342/2 = .171
θ₂ = 9.8 °
Answer: 2934.75 Joules
Explanation:
Potential energy can be defined as energy possessed by an object or body due to its position.
Mathematically, potential energy is given by the formula;
<em>P.E = mgh</em>
Where P.E represents potential energy measured in Joules.
m represents the mass of an object.
g represents acceleration due to gravity measured in meters per second square.
h represents the height measured in meters.
Given the following data;
Weight =645
Height = 4.55
<em>P.E = mgh</em>
But we know that weight = mg = 645N
Substituting into the equation, we have;
<em>P.E = 645 • 4.55</em>
<em>P.E = 2934.75J</em>
Potential energy, P.E = 2934.75 Joules.
Answer:
8.4 V
Explanation:
induced emf, e1 = 5.8 V
Magnetic field, B1 = 0.38 T
magnetic field, B2 = 0.55 T
induced emf, e2 = ?
As we know that the induced emf is directly proportional to the magnetic field strength.
When the other parameters remains constant then
e2 = 8.4 V
Thus, the induced emf is 8.4 V.
A stable air mass is most likely to have POOR SURFACE VISIBILITY.
Stable air mass refers to those air mass that have marked stability in their lower layers. The characteristics of stable air mass include the following: cloud cover, smooth air, uninterrupted precipitation and low visibility.<span />
Answer:
a). M = 20.392 kg
b). am = 0.56 
(block), aM = 0.28 (bucket)
Explanation:
a). We got N = mg cos θ,
f = 
= 
If the block is ready to slide,
T = mg sin θ + f
T = mg sin θ + .....(i)
2T = Mg ..........(ii)
Putting (ii) in (i), we get
M = 20.392 kg
b). 
.............(iii)
Here, l = total string length
Differentiating equation (iii) double time w.r.t t, l, h and h' are constants, so
.....................(iv)
We got, N = mg cos θ
∴ 
................(v)
Mg - 2T = M
(from equation (iv))
.....................(vi)
Putting (vi) in equation (v),
![$\frac{g\left[\frac{M}{2}-m \sin \theta-\mu_K m \cos \theta\right]}{(\frac{M}{4}+m)}=a_m$](https://tex.z-dn.net/?f=%24%5Cfrac%7Bg%5Cleft%5B%5Cfrac%7BM%7D%7B2%7D-m%20%5Csin%20%5Ctheta-%5Cmu_K%20m%20%5Ccos%20%5Ctheta%5Cright%5D%7D%7B%28%5Cfrac%7BM%7D%7B4%7D%2Bm%29%7D%3Da_m%24)
![$\frac{9.8\left[\frac{20.392}{2}-10(\sin 30+0.5 \cos 30)\right]}{(\frac{20.392}{4}+10)}=a_m$](https://tex.z-dn.net/?f=%24%5Cfrac%7B9.8%5Cleft%5B%5Cfrac%7B20.392%7D%7B2%7D-10%28%5Csin%2030%2B0.5%20%5Ccos%2030%29%5Cright%5D%7D%7B%28%5Cfrac%7B20.392%7D%7B4%7D%2B10%29%7D%3Da_m%24)
Using equation (iv), we get,
